Ex 5.1 ,1 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Feb. 14, 2025 by Teachoo
Last updated at Feb. 14, 2025 by Teachoo
Ex 5.1, 1 Prove that the function π (π₯) = 5π₯ β 3 is continuous at π₯ = 0, at π₯ = β3 and at π₯ = 5 Given π(π₯)= 5π₯ β3 At π=π f(x) is continuous at x = 0 if (π₯π’π¦)β¬(π±βπ) π(π) = π(π) (π₯π’π¦)β¬(π±βπ) π(π) "= " limβ¬(xβ0) " "(5π₯β3) Putting x = 0 = 5(0) β 3 = β3 π(π) = 5(0) β 3 = 0 β 3 = β3 Since L.H.S = R.H.S Hence, f is continuous at π = π At x = β3 f(x) is continuous at x = β3 if ( lim)β¬(xββ3) π(π₯)= π(β3) Since, L.H.S = R.H.S Hence, f is continuous at π =β3 (π₯π’π¦)β¬(π±βπ) π(π) "= " limβ¬(xβ3) " "(5π₯β3) Putting x = β3 = 5(β3) β 3 = β18 π(βπ) = 5(β3) β 3 = β15 β 3 = β18 At π =π f(x) is continuous at x = 5 if ( lim)β¬(xβ5) π(π₯)= π(5) Since, L.H.S = R.H.S Hence, f is continuous at x = 5 Thus, the function is continuous at x = 0, x = β3, x = 5 (πππ)β¬(πβπ) π(π) "= " (πππ)β¬(π₯β5) " "(5π₯β3) Putting x = 5 = 5(5) β 3 = 22 π(π) = 5(5) β 3 = 25 β 3 = 22
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo